Chem Lecture Test 1 IPT

Chem  Lecture  T1  15  JD  TY  N   Memory  Work       What  to  memorise   Isotopes  

Ionisation  energy  

Metallic  Bonding   Ionic  Bonding   Covalent  Bonds   Explaining  why  dative  bonds  form  

Polarity  

Definition   Have  same  proton  number  but  different   nucleon  number     Same  chemical  properties,  but  different   physics  properties     Atoms  of  an  element   First  Ionisation  energy  is  the  amount  of   energy  required  to  remove  1  mole  of   electrons  from  1  mole  of  gaseous  atoms  to   form  1  mole  of  unipositive  gaseous  ions.     Second  ionisation  energy  is  the  amount  of   energy  required  to  remove  1  mole  of   electrons  from  1  mole  of  unipositive   gaseous  ions  to  form  1  mole  of  gaseous   ions  with  double  positive  charge.   Electrostatic  force  of  attraction  between   metal  cations  and  sea  of  delocalised   electrons   Electrostatic  force  of  attraction  between   oppositely  charged  ions   Electrostatic  force  of  attraction  between   nucleus  of  the  two  bonded  atoms  ahd   shared  pair  of  electrons   Because  one  of  the  molecular  has  lone   pairs  and  can  act  as  an  electron  donor   while  the  other  molecular  has  only  1/2/3   electron  pairs,  has  an  empty  orbital  in   valence  shell  and  can  accept  the  donated   electrons  to  achieve  octet.   Separation  of  Charge,  unequal  sharing  of   charge,  dipole  moments  

  Explaining  solubility/interaction  between  ionic  and  covalent  substances     Theory   Substantiation       Covalent   Covalent       Energy  released  from  Solute  solvent   Need  to  IDENTIFY  the  different  types  of   interaction    greater  than/smaller  than   interaction  in  solute  solute,  solvent  solvent,   energy  required  to  break  solute  solute  and   solute  solvent  before  elaborating.   solvent  solvent  bonds    

Chem  Lecture  T1  15  JD  TY  N    

Then  proceed  to  explain  which  bond  is  the   stronger  one  and  how  it  leads  to  the   solubility.  

Ionic  Compounds         Need  to  IDENTIFY  the  different  types  of   Energy  released  from  ion-­‐dipole   interaction  in  solute  solute,  solvent  solvent,   interactions  >/<  energy  needed  to  break   solute  solvent  before  elaborating.   hydrogen  bonds  of  water  and  solute  solute     bonds,  detaches  ion  from  surface   Then  proceed  to  explain  which  bond  is  the   stronger  one  and  how  it  leads  to  the   solubility.    

Sketching  Important  Trends  

  Trend   Atomic   Radius   Across  the   Period  

Sketch+Explain  

1.   Atomic  radius  decreases  down  the  period  as     -­‐   Nuclear  charge  increases  due  to  increasing  proton  number   -­‐   Shielding  effect  remains  relatively  constant   -­‐   Effective  Nuclear  Charge  increases   -­‐   Electrostatic  forces  of  attraction  between  ____  and  _____   increases  

Chem  Lecture  T1  15  JD  TY  N   -­‐   More  energy  required  to  overcome  

                                   

2.   Ionic  radius  generally  decreases  down  the  period,  except  for  huge   jump  from  cations  to  anions   -­‐   Nuclear  charge  increases   -­‐   Shielding  effect  remains  relatively  constant   -­‐   Effective  Nuclear  Charge  increases   -­‐   Electrostatic  forces  of  attraction  between  ____  and  _____   increases   -­‐   Sharp  jump  from  cations  to  anions  due  to  additional  quantum   shell  of  electrons  (3s  and  3p)  

3.   Trends  in  successive  I.E   -­‐   Small  increases  when  moving  from  p  orbital  to  s  as  s  orbitals  at   higher  energy  level  as  they  are  CLOSER  to  the  nucleus   -­‐   Large  jump  when  next  electron  is  removed  from  an  inner   quantum  shell  (look  at  the  number  of  valence  electrons)  

Chem  Lecture  T1  15  JD  TY  N                            

 

  4.   Trends  in  IE  across  period   -­‐   Nuclear  charge  increases  across  period   -­‐   Shielding  effect  remains  relatively  constant   -­‐   ENC  increases   -­‐   Electrostatic  forces  of  attraction  between  ___  and  ___   increases   -­‐   More  energy  needed  to  remove  outermost  electron   Some  anomalies   -­‐   Small  dip  from  Mg  to  Al  as  3p  subshell  of  al  at  higher  energy   level  than  3s  of  Mg,  weaker  attraction  between  nucleus  and   outermost  electron,  less  energy  required.   -­‐   Inter-­‐electronic  repulsion  for  paired  electrons  in  3p  subshell  of  S,   less  energy  required  to  remove  the  more  unstable  electrons.    

 

Deducing  strength  of  Bonds     Type  of  bond   Metallic  

Ionic  

Covalent  

Explanation   -­‐Compare  number  of  valence  electrons   contributed  PER  METAL  ATOM   (Stronger/weaker  electrostatic  forces  of   attraction)     -­‐Charge,  radius  and  charge  density.  Larger   the  charge  and  smaller  the  radius,  ,the   stronger  the  bond.   Lattice  Energy  (Charge  of  cation,  anion,   radius  of  cation,  anion)       Remember  to  break  cations  and  anions  up   separately  to  compare  and  analyze.   Polarity,  size  and  number  of  bonds  (The   smaller  the  atom,  the  more  effective  the   orbital  overlap,  the  closer  to  the  nucleus.  

Chem  Lecture  T1  15  JD  TY  N   The  more  polar,  the  shorter  the  bond   length.  The  more  number  of  bonds,  the   stronger  the  attraction  between  nucleus   and  shared  electrons.)     Explaining  Chemical/Physical  Properties     Type   Properties   Metals   1.   High  Melting  and  Boiling  Points   -­‐   Exist  as  giant  metallic  lattice,   strong  electrostatic  forces  of   attraction  between  metal   cations  and  sea  of  delocalised   electrons,  more  energy  needed   to  break..   2.   Good  electrical  conductivity   -­‐   Sea  of  delocalised  electrons  to   act  as  mobile  charge  carriers   3.   Good  heat  conductors   -­‐   Heat  energy  picked  up  by   electrons  to  vibrate  faster   (kinetic  energy)   4.   Shiny  surface   -­‐   When  photon  strike  surface,   osicillating  electric  field  causes   electrons  on  surface  to  oscillate.   Photon  bounces  off  without  loss   of  momentum.   5.   Hardness   -­‐   Strong  metallic  bond   6.   Malleable  and  ductile   -­‐   Layers  of  ions  can  slide  over  one   another  into  new  position.  Sea   of  delocalised  electrons  reduces   repulsions  between  cations,  so   do  not  break.   7.   Alloy   -­‐   Disrupt  orderly  arrangement  of   metal  atoms   Ionic  Compounds     1.   High  Melting  and  Boiling  Points   -­‐   Exist  as  giant  metallic  lattice,   strong  electrostatic  forces  of   attraction  between  metal   cations  and  sea  of  delocalised   electrons,  more  energy  needed   to  break..  

Chem  Lecture  T1  15  JD  TY  N  

Covalent  

2.   Good  Electrical  conductors  in   molten/aqueous  state   -­‐   Ions  in  solid  state  are  held  in  a   fixed  lattice,  but  when   dissociated  can  act  as  mobile   charge  carriers   3.   Hard,  rigid  but  brittle   -­‐   Hard  force  may  cause  ions  of   like  charges  to  become  next  to   each  other,  causing  repulsion   and  lattice  to  shatter.   1.   Low  melting  and  boiling  point  for   simple  molecular  structures  and   high  for  Giant  covalent  structures   -­‐   SMS  molecules  held  by  weak   VDW  forces   -­‐   GCS  molecules  held  by  strong   covalent  bonds   2.   Most  do  not  conduct  electricity   except  for  Graphite   -­‐   As  electrons  are  held  in  covalent   bonds/nucleus     -­‐   As  for  graphite,  each  carbon  is   only  covalently  bonded  to  three   others,  there  is  a  spare  electron   to  act  as  mobile  charge  carrier       Drawing  Sigma/PI  bonds  

  Type   Drawing   of   Bond   Sigm a   Bond  

 

Chem  Lecture  T1  15  JD  TY  N   Pi   Bond  

 

1s,   2s,  3s  

   

Note:  2s  orbital  larger  than  1s,  3s  larger  than  2s     Explaining  difference  between  bond  angles     Theory   Explanation   1.   Compare  number  of  electron  pairs   If  electron  pairs  is  different,  then   first   immediately  use  VSEPR  to  explain       2.   Compare  number  of  lone  pairs  if   Lone  pair  lone  pair  repulsion  is  the   electron  pairs  is  the  same   strongest,  so  it  will  push  bond  pairs     closer.   3.   If  all  are  the  same,  use  polarity.  A     more  polar  molecular  will  have   Remember  the  sequence=  LP-­‐LP     slightly  differing  bond  angles  than  a   repulsion>  LP-­‐BP  Repulsion  >  BP-­‐BP   less  polar  one.     repulsion.  Electrons  in  a  more  polar   molecule  will  be  attracted  to  __,   leading  to  decrease  in  BP-­‐BP   repulsion  (depends)  and  increase  in   LP-­‐BP  repulsion.            

 

Determining  Difference  Between  Melting/Boiling  Points  

  Key  

Possible  ways  of  Answering  Required  

Chem  Lecture  T1  15  JD  TY  N   Remember  that  melting/boiling  points  are   always  due  to  intermolecular  forces  of   attraction.     Ionic  compounds:  Ions   Metallic  compounds:  Ions  and  Delocalised   electrons   Covalent  compounds:  Molecules  

 

Check  for  Differences  in  type  of   bonding   1.   Comparing  the  structure    (Ionic  VS   Covalent  VS  metallic  compounds)     -­‐   Giant  ionic  lattice,  simple   molecular  structure,  giant   molecular  structure,  giant   metallic  lattice   Check  for  Hydrogen  Bonds  

  2.   Hydrogen  bonds  are  strong   -­‐   Requires  FON  with  lone   pair   -­‐   H  bonded  to  FON     Confirm  the  Intermolecular  Force     3.   Polar  or  Non-­‐Polar   -­‐   Non-­‐polar  compounds  have   dispersion  forces,  which  are   weaker  than  the  permanent   dipole  permanent  dipole   interactions  in  polar  compounds     Non  Polar  Compounds   Check  for  difference  in  Mr     4.   If  both  molecules  are  non-­‐ polar,  compare  Mr   -­‐   Larger  Mr  means   larger  electron  cloud,   more  polarisable,   more  formation  of   induced  and   instantaneous  dipoles,   stronger  dispersion   forces     Check  for  Branches     5.   If  both  molecules  are  non-­‐ polar,  compare  the   surface  area   -­‐   A  straight  chain   compound  has  larger   points  of  comtact  that  

Chem  Lecture  T1  15  JD  TY  N   dispersion  forces  can   act.  Therefore,   stronger  dispersion   forces.      

Polar  Compounds   Difference  in  Polarity     6.   Increase  in  polarity  leads   to  shorter  bond  length   -­‐   More  polar  means   shorter  bond  length,   electron  more   attracted  to  nucleus,   more  energy  needed   to  break  bonds